Bar Bending Schedule For Circular Slab [ BBS ]

Suppose we have a circular slab having a diameter 10 ft. The main and distribution bars were spaced 6 inches both ways. So calculate the number of bars?

Given data:

Diameter of slab = 10 feet

Main Bar = #10 @ 6 inch c/c

Distribution bar = #10 @ 6 inch c/c

Solution:

No of bars:

Formula

= diameter of bar / center to center distance - 1

= 10/0.5 - 1

= 19 bars

One bar is basically in mid of slab which length is almost 10 feet

Formula:

= 2 * (R)m² - ( c/c )²m²

#where

R = Radius of circle c/c = center to center distance (it will change as we more to next bar calculation)

1st = 2 * V((5)² - (0.5)²) = 9.94 ft

2nd = 2 * V((5)² - (1)²) = 9.79 ft

3rd = 2 * V((5)² - (1.5)²) = 9.53 ft

4th = 2 * V((5)² - (2)²) = 9.16 ft

5th = 2 * V((5)² - (2.5)²) = 8.66 ft

6th = 2 * V((5)² - (3.0)²) = 8 ft

7th = 2 * V((5)² - (3.5)²) = 7.14 ft

8th = 2 * V((5)² - (4)²) = 6 ft

9th = 2 * V((5)² - (4.5)²) = 4.35 ft

So two nos of bar is cut out for the same Length of 1st 2nd and 3rd bar Length because the slab is circular in shape.